Problem: Simplify; express your answer in exponential form. Assume $x\neq 0, p\neq 0$. $\dfrac{{(x^{-2})^{-2}}}{{(x^{-3}p^{4})^{2}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${x^{-2}}$ to the exponent ${-2}$ . Now ${-2 \times -2 = 4}$ , so ${(x^{-2})^{-2} = x^{4}}$ In the denominator, we can use the distributive property of exponents. ${(x^{-3}p^{4})^{2} = (x^{-3})^{2}(p^{4})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(x^{-2})^{-2}}}{{(x^{-3}p^{4})^{2}}} = \dfrac{{x^{4}}}{{x^{-6}p^{8}}}$ Break up the equation by variable and simplify. $\dfrac{{x^{4}}}{{x^{-6}p^{8}}} = \dfrac{{x^{4}}}{{x^{-6}}} \cdot \dfrac{{1}}{{p^{8}}} = x^{{4} - {(-6)}} \cdot p^{- {8}} = x^{10}p^{-8}$.